Question

Maximize \(3 a x^{3}-b x^{3}-\frac{2 b^{2} a}{3 c} x+a^{2} b\) using Hudde's rule. (This example is taken from Hudde's De maximis et minimis.)

Short answer

Question: Using Hudde's rule, maximize the function \(3ax^3 - bx^3 - \frac{2b^2a}{3c}x + a^2b\). Answer: To maximize the function, first find the critical points by setting the derivative equal to 0: \(x = \pm \sqrt{\frac{2b^2a}{3c(9a - 3b)}} \). Check which of these values results in a local maximum by evaluating the second derivative at these x-values. The maximum turning point will be the x-value that gives a negative result when evaluated at the second derivative.

Step-by-step solution

Understand Hudde's Rule

Hudde's rule is a method used to find the maximum or minimum value of a function. The rule states that if a function f(x) has a turning point at x=c, then the derivative of f(x), f'(x) must be equal to 0 at x=c. So, in order to find the critical or turning points of our function, we need to find its first derivative and set it to 0.

Find the First Derivative

The function we're working with is \(3ax^3 - bx^3 - \frac{2b^2a}{3c}x + a^2b\). To find its first derivative, we will differentiate this function with respect to x. Apply the power rule and differentiate each term: f'(x) = \( \frac{d}{dx}(3ax^3) - \frac{d}{dx}(bx^3) - \frac{2b^2a}{3c} \cdot \frac{d}{dx}(x) + \frac{d}{dx}(a^2b) \) Finding the derivatives of each term individually, we get: f'(x) = \( 9ax^2 - 3bx^2 - \frac{2b^2a}{3c} \) Now that we have the first derivative of the function, we can proceed to find critical points by setting f'(x) equal to 0.

Determine the Critical Points

To find the critical points, set the first derivative to 0 and solve for x: \( 0 = 9ax^2 - 3bx^2 - \frac{2b^2a}{3c} \) Since we are only interested in the values of x for which the function has turning points, we'll collect the terms containing x and factor them. \( 0 = x^2(9a - 3b) - \frac{2b^2a}{3c} \)

Solve for x

Now we need to solve this equation to find the value of x: \( x^2(9a - 3b) = \frac{2b^2a}{3c} \) Divide both sides by (9a - 3b): \( x^2 = \frac{2b^2a}{3c(9a - 3b)} \) Taking the square root of both sides: \( x = \pm \sqrt{\frac{2b^2a}{3c(9a - 3b)}} \) The two values of x represent the turning points of the function. To maximize the function, we need to determine which of these values results in a maximum.

Determine the Maximum

To check if a value of x corresponds to a maximum or minimum, evaluate the second derivative at that x-value. If it is negative, then it is a local maximum. If it is positive, then it is a local minimum. Find the second derivative by taking the derivative of f'(x): f''(x) = \(\frac{d}{dx}(9ax^2 - 3bx^2 - \frac{2b^2a}{3c})\) = \( 18ax - 6bx \) Evaluate f''(x) at both the positive and negative solution for x: \( f''(\sqrt{\frac{2b^2a}{3c(9a - 3b)}}) = 18a\sqrt{\frac{2b^2a}{3c(9a - 3b)}} - 6b\sqrt{\frac{2b^2a}{3c(9a - 3b)}} \) Since the values of a, b, and c are non-zero, the maximum turning point will correspond to whichever value of x gives a negative result when evaluated at f''(x). To maximize the given function, consider the obtained critical points and evaluate them to determine the local maximum.

Key concepts

Maxima and Minima

Maxima and minima refer to the highest and lowest points of a function, respectively. These are important concepts in calculus because they help determine where a function reaches its peaks and valleys.

To find them, you first look for critical points using the derivative of the function. This involves setting the first derivative equal to zero and solving for the variable. Critical points can be either maxima, minima, or saddle points (points that are neither).

But how can you tell which is which? This is where the second derivative comes into play. By evaluating the second derivative at the critical points:

• If it is positive, the critical point is a local minimum.
• If it's negative, you have a local maximum.
• If the second derivative is zero, more investigation will be needed, potentially reverting to higher-order derivatives.

Understanding maxima and minima can help optimize various real-world scenarios, from maximizing profits to minimizing costs. It's a powerful tool for decision-making and is fundamental in mathematical analysis.

Calculus

Calculus is a branch of mathematics focusing on rates of change (differential calculus) and accumulation of quantities (integral calculus). It plays a pivotal role in many fields such as physics, engineering, economics, and beyond.

The core idea of calculus is to study how things evolve over time. Functions describe these changing entities, and calculus provides tools to analyze behaviors of these functions. The two main components of calculus are:

• **Differentiation:** It helps us understand the rate at which something changes. Think of it as assessing the slope of a curve, indicating how steep or gentle a function is at any particular point.
• **Integration:** While differentiation provides the rate of change, integration tells us about the underlying quantity. It's akin to finding the area under a curve, showing the full effect of accumulated change.

Calculus forms the backbone of many complex theories and is indispensable in modeling real-life situations, lending insight into systems that are constantly evolving.

Derivative

A derivative primarily measures how a function changes as its input changes. It's the core tool for understanding and identifying maxima and minima. The derivative is calculated using differentiation, which is foundational in calculus.

When you find the derivative of a function, you're essentially looking for the slope of the tangent line to the function at any point. For instance:

• A positive derivative means the function is increasing.
• A negative derivative suggests the function is decreasing.
• A derivative of zero indicates a potential maximum or minimum — critical points where the function might change direction.

Moreover, the first derivative is used to find critical points, which are candidates for being maxima or minima. When you calculate a derivative, you're employing rules such as the power rule, product rule, quotient rule, and chain rule to simplify the process.

These rules make it easier to derive even complex functions and are standard tools in any calculus toolbox. Understanding and effectively utilizing derivatives is a substantial step toward mastering calculus.